let f(x) =∫_0 ^2 (√(x+t^2 ))dt with x≥0 1) calculate f(x) 2)calculate g(x) =∫_0 ^2 (dt/(√(x+t^2 ))) 3)find the value[of ∫_0 ^2 (√(4+t^2 ))dt and ∫_0 ^2 (dt/(√(3+t^2 ))) 4) give g^′ (x) at form of integral.


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Question Number 66801 by mathmax by abdo last updated on 19/Aug/19

$l e t f (x) = \int_{0}^{2} \sqrt{x + t^{2}} d t w i t h x ⩾ 0$ $1) c a l c u l a t e f (x)$ $2) c a l c u l a t e g (x) = \int_{0}^{2} \frac{d t}{\sqrt{x + t^{2}}}$ $3) f i n d t h e v a l u e [o f \int_{0}^{2} \sqrt{4 + t^{2}} d t a n d \int_{0}^{2} \frac{d t}{\sqrt{3 + t^{2}}}$ $4) g i v e g^{^{'}} (x) a t f o r m o f i n t e g r a l .$
Commented by mathmax by abdo last updated on 21/Aug/19
$1) f(x)=∫_0 ^2 (√(x+t^2 ))dt changement t =(√x)u give f(x)=∫_0 ^(2/(√x)) (√x)(√(1+u^2 ))(√x)du =x ∫_0 ^(2/(√x)) (√(1+u^2 ))du changement u=shz give f(x) =x ∫_0 ^(argsh((2/(√x)))) ch(z)ch(z)ez =(x/2) ∫_0 ^(argsh((2/((√)x)))) (ch(2z)+1)dz =(x/4)[sh(2z)]_0 ^(ln((2/(√x))+(√(1+(4/x))))) +(x/2)ln((2/(√x))+(√(1+(4/x)))) =(x/8)[e^(2z) −e^(−2z) ]_0 ^(ln(((2+(√(x+4)))/((√x) )))) +(x/2)ln(((2+(√(x+4)))/(√x))) ⇒ f(x)=(x/8){(((2+(√(x+4)))/(√x)))^2 −(1/((((2+(√(x+4)))/(√x)))^2 ))}+(x/2)ln(((2+(√(x+4)))/(√x))) 2)we have f^′ (x)= ∫_0 ^2 (1/(2(√(x+t^2 )))) dt =(1/2)g(x) ⇒g(x) =2f^′ (x) rest to calculate f^′ (x) ..be continued...$
$1) f (x) = \int_{0}^{2} \sqrt{x + t^{2}} d t c h a n g e m e n t t = \sqrt{x} u g i v e$ $f (x) = \int_{0}^{\frac{2}{\sqrt{x}}} \sqrt{x} \sqrt{1 + u^{2}} \sqrt{x} d u = x \int_{0}^{\frac{2}{\sqrt{x}}} \sqrt{1 + u^{2}} d u c h a n g e m e n t u = s h z$ $g i v e f (x) = x \int_{0}^{a r g s h (\frac{2}{\sqrt{x}})} c h (z) c h (z) e z = \frac{x}{2} \int_{0}^{a r g s h (\frac{2}{\sqrt{} x})} (c h (2 z) + 1) d z$ $= \frac{x}{4} {[s h (2 z)]}_{0}^{l n (\frac{2}{\sqrt{x}} + \sqrt{1 + \frac{4}{x}})} + \frac{x}{2} l n (\frac{2}{\sqrt{x}} + \sqrt{1 + \frac{4}{x}})$ $= \frac{x}{8} {[e^{2 z} - e^{- 2 z}]}_{0}^{l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}})} + \frac{x}{2} l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}}) \Rightarrow$ $f (x) = \frac{x}{8} {{(\frac{2 + \sqrt{x + 4}}{\sqrt{x}})}^{2} - \frac{1}{{(\frac{2 + \sqrt{x + 4}}{\sqrt{x}})}^{2}}} + \frac{x}{2} l n (\frac{2 + \sqrt{x + 4}}{\sqrt{x}})$ $2) w e h a v e f^{^{'}} (x) = \int_{0}^{2} \frac{1}{2 \sqrt{x + t^{2}}} d t = \frac{1}{2} g (x) \Rightarrow g (x) = 2 f^{^{'}} (x)$ $r e s t t o c a l c u l a t e f^{^{'}} (x) . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 21/Aug/19
$3) we have ∫_0 ^2 (√(4+t^2 ))dt =f(4) =(1/2){(((2+2(√2))/2))^2 −(1/((((2+2(√2))/2))^2 )) +2ln(((2+2(√2))/2)) =(1/2){ (1+(√2))^2 −(1/((1+(√2))^2 ))} +2ln(1+(√2)) ∫_0 ^2 (dt/(√(3+t^2 ))) =_(t=(√3)u) ∫_0 ^(2/(√3)) (((√3)du)/((√3)(√(1+u^2 )))) =∫_0 ^(2/(√3)) (du/(√(1+u^2 ))) =[ln(u+(√(1+u^2 )))]_0 ^(2/(√3)) =ln((2/(√3)) +(√(1+(4/3)))) =ln(((2+(√7))/(√3)))$
$3) w e h a v e \int_{0}^{2} \sqrt{4 + t^{2}} d t = f (4) = \frac{1}{2} {{(\frac{2 + 2 \sqrt{2}}{2})}^{2} - \frac{1}{{(\frac{2 + 2 \sqrt{2}}{2})}^{2}}$ $+ 2 l n (\frac{2 + 2 \sqrt{2}}{2}) = \frac{1}{2} {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}} + 2 l n (1 + \sqrt{2})$ $\int_{0}^{2} \frac{d t}{\sqrt{3 + t^{2}}} =_{t = \sqrt{3} u} \int_{0}^{\frac{2}{\sqrt{3}}} \frac{\sqrt{3} d u}{\sqrt{3} \sqrt{1 + u^{2}}} = \int_{0}^{\frac{2}{\sqrt{3}}} \frac{d u}{\sqrt{1 + u^{2}}}$ $= {[l n (u + \sqrt{1 + u^{2}})]}_{0}^{\frac{2}{\sqrt{3}}} = l n (\frac{2}{\sqrt{3}} + \sqrt{1 + \frac{4}{3}}) = l n (\frac{2 + \sqrt{7}}{\sqrt{3}})$
Commented by mathmax by abdo last updated on 21/Aug/19

$4) w e h a v e g (x) = \int_{0}^{2} {(x + t^{2})}^{- \frac{1}{2}} d t \Rightarrow g^{^{'}} (x) = \int_{0}^{2} - \frac{1}{2} {(x + t^{2})}^{- \frac{3}{2}} d t$ $= - \frac{1}{2} \int_{0}^{2} \frac{d t}{(x + t^{2}) \sqrt{x + t^{2}}} .$
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